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lst={}; Do[i=3^n/n^3; If[IntegerQ[i], AppendTo[lst, n]], {n, 6!}]; lst
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a(n) = 3^(3^n)/(3^n)^3.
a(n) = 3^A107583(n). [_R. J. Mathar, _, Oct 07 2009]
a(n=0) = 3^1/1^3 = 3. a(n=1) = 3^3/3^3 = 1. a(n=2) = 3^9/9^3 = 27.
approved
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Table[3^(3^n)/(3^n)^3, {n, 0, 6}] (* Harvey P. Dale, Sep 17 2019 *)
approved
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_Vladimir Joseph Stephan Orlovsky (4vladimir(AT)gmail.com), _, Sep 22 2009
Definition simplified by _R. J. Mathar (mathar(AT)strw.leidenuniv.nl), _, Oct 07 2009